Ideal gas: Energy: Difference between revisions
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The energy of the [[ideal gas]] is given by (Hill Eq. 4-16) | The energy of the [[ideal gas]] is given by (Hill Eq. 4-16) | ||
:<math>E = -T^2 \left. \frac{\partial (A/T)}{\partial T} \right\vert_{V,N} = kT^2 \left. \frac{\partial \ln Q}{\partial T} \right\vert_{V,N}= NkT^2 \frac{d \ln T^{3/2}}{dT} = \frac{3}{2} NkT</math> | :<math>E = -T^2 \left. \frac{\partial (A/T)}{\partial T} \right\vert_{V,N} = kT^2 \left. \frac{\partial \ln Q}{\partial T} \right\vert_{V,N}= NkT^2 \frac{d \ln T^{3/2}}{dT} = \frac{3}{2} NkT \equiv \frac{3}{2} RT </math> | ||
where <math>R</math> is the [[molar gas constant]]. | |||
This energy is all ''kinetic energy'', <math>1/2 kT</math> per [[degree of freedom]], by [[equipartition]]. This is because there are no intermolecular forces, thus no potential energy. This result is valid only for a monoatomic ideal gas. The general expression would be | |||
:<math>E = \frac{n}{2} NkT = \frac{n}{2} RT, </math> | |||
where <math>n</math> is the number of degrees of freedom. This number is 3 for atoms; if would be 6 in principle for diatomic molecules, but in normal conditions 5 is a very good approximation since vibrations are "frozen" (as explained in the entry about [[degree of freedom | degrees of freedom]].) | |||
==References== | ==References== | ||
#Terrell L. Hill "An Introduction to Statistical Thermodynamics" 2nd Ed. Dover (1962) | #Terrell L. Hill "An Introduction to Statistical Thermodynamics" 2nd Ed. Dover (1962) | ||
[[category: ideal gas]] | [[category: ideal gas]] | ||
Latest revision as of 15:44, 12 December 2008
The energy of the ideal gas is given by (Hill Eq. 4-16)
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = -T^2 \left. \frac{\partial (A/T)}{\partial T} \right\vert_{V,N} = kT^2 \left. \frac{\partial \ln Q}{\partial T} \right\vert_{V,N}= NkT^2 \frac{d \ln T^{3/2}}{dT} = \frac{3}{2} NkT \equiv \frac{3}{2} RT }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} is the molar gas constant. This energy is all kinetic energy, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/2 kT} per degree of freedom, by equipartition. This is because there are no intermolecular forces, thus no potential energy. This result is valid only for a monoatomic ideal gas. The general expression would be
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = \frac{n}{2} NkT = \frac{n}{2} RT, }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is the number of degrees of freedom. This number is 3 for atoms; if would be 6 in principle for diatomic molecules, but in normal conditions 5 is a very good approximation since vibrations are "frozen" (as explained in the entry about degrees of freedom.)
References[edit]
- Terrell L. Hill "An Introduction to Statistical Thermodynamics" 2nd Ed. Dover (1962)